User blog:Rgetar/Regular and singular ordinals
I was trying to extend my Veblen-like functions beyond Ωω, and there was a problem: for Veblen-like function φαβ(X) should be cof(X) ≤ Ωα + 1 (cof(X), that is cofinality of X, is minimal length of increasing sequence of ordinals such as its supremum is X). But what if cof(X) > Ωα + 1? Then we can use φαβ(X) = φαγ(φγβ(X)) For example, φ26100(Ω88) cof(Ω88) = Ω88 > Ω26 + 1 But let γ = 87, then φ26100(Ω88) = φ2687(φ87100(Ω88)) cof(Ω88) = Ω88 = Ω87 + 1 cof(φ87100(Ω88)) = ω < Ω26 + 1 So everything is fine. Then let β = ω φ0ω(0) cof(0) = 0 < Ω0 + 1 = Ω ok. φ0ω(1) cof(1) = 1 < Ω ok. φ0ω(ω) cof(ω) = ω < Ω ok. φ0ω(Ω) cof(Ω) = Ω = Ω ok. φ0ω(Ω2) cof(Ω2) = Ω2 > Ω Not ok. Then φ0ω(Ω2) = φ01(φ1ω(Ω2)) ok. Similarly φ0ω(Ω3) = φ02(φ2ω(Ω3)) φ0ω(Ω4) = φ03(φ3ω(Ω4)) φ0ω(Ω5) = φ04(φ4ω(Ω5)) ... But for φ0ω(Ωω) we need γ φ0ω(Ωω) = φ0γ(φγω(Ωω)) such as γ + 1 = ω? However, this is impossible, since ω is not successor ordinal... Then I thought: "Wait, cof(Ωω) is not Ωω, it is ω, since Ωω is supremum of ω, Ω1, Ω2, Ω3, Ω4, Ω5..." Interestingly, cof(ω) = ω cof(Ω) = Ω cof(Ω2) = Ω2 cof(Ω3) = Ω3 cof(Ω4) = Ω4 cof(Ω5) = Ω5 ... cof(Ωω) = ω ≠ Ωω Apparently, cof(Ωω + 1) = Ωω + 1 cof(Ωω + 2) = Ωω + 2 cof(Ωω + 3) = Ωω + 3 ... cof(Ωω2) = ω (since it is supremum of Ωω, Ωω + 1, Ωω + 2, Ωω + 3, Ωω + 4, Ωω + 5...) cof(ΩΩ) = Ω cof(ΩΩ + 1) = ΩΩ + 1 cof(ΩΩ + ω) = ω cof(ΩΩ2) = Ω2 cof(ΩΩ2 + 1) = ΩΩ2 + 1 cof(ΩΩ3) = Ω3 cof(ΩΩω) = ω cof(ΩΩΩ) = Ω cof(ΩΩΩω) = ω cof(ΩΩΩΩ) = Ω cof(ΩΩΩΩΩ...) = ω (since it is supremum of ω, Ωω, ΩΩω, ΩΩΩω, ΩΩΩΩω, ...) I proved that there are no ordinals of cofinality Ωω. Proof: let cof(α) = Ωω. Then α is supremum of increasing sequence αn, where n are all ordinals less than Ωω. Remove from this sequence all elements except αΩβ, where β are natural numbers. We get αω, αΩ, αΩ2, αΩ3, αΩ4, αΩ5, ... α is also supremum of this new sequence of length ω, so cof(α) = ω. And there are no ordinals of cofinality Ωω2. Proof is similar, but with α+ β instead of αΩβ. etc. So, since there are no ordinals of cofinality Ωω, the problem with φαω(X) is solved, since cof(X) < Ωω, for example, φ0ω(Ωω) cof(Ωω) = ω < Ω We even do not need γ. And similarly for φαω2(X), etc. So, apparently, if α is 0 or successor ordinal, then cof(Ωα) = Ωα and if α is limit ordinal, then possibly not. I was wondering, if opposite true, that is if cof(Ωα) ≠ Ωα for all limit α. I needed to check it all out. I read Wikipedia article Cofinality, and it turned out that this is really so: ordinals α such as cof(Ωα) = Ωα are called "regular", and all other ordinals are called "singular". Most ordinals are singular. Ordinals Ωα for zero or successor α are regular. Ordinals Ωα for limit α are usually singular. But it was still unclear: are there regular Ωα with limit α? Or all Ωα with limit α are singular? I searched "cofinality" at Googology Wiki and read article Cardinal. And the answer is "maybe". Regular Ωα with limit α are called "inaccessible cardinals". Least inaccessible cardinal is called I. Category:Blog posts